Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $a = \dfrac{-4t - 20}{t - 10} \times \dfrac{-4t^2 + 24t + 160}{t^2 + 11t + 30} $
Answer: First factor out any common factors. $a = \dfrac{-4(t + 5)}{t - 10} \times \dfrac{-4(t^2 - 6t - 40)}{t^2 + 11t + 30} $ Then factor the quadratic expressions. $a = \dfrac {-4(t + 5)} {t - 10} \times \dfrac {-4(t - 10)(t + 4)} {(t + 5)(t + 6)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {-4(t + 5) \times -4(t - 10)(t + 4) } {(t - 10) \times (t + 5)(t + 6) } $ $a = \dfrac {16(t - 10)(t + 4)(t + 5)} {(t + 5)(t + 6)(t - 10)} $ Notice that $(t + 5)$ and $(t - 10)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {16(t - 10)(t + 4)\cancel{(t + 5)}} {\cancel{(t + 5)}(t + 6)(t - 10)} $ We are dividing by $t + 5$ , so $t + 5 \neq 0$ Therefore, $t \neq -5$ $a = \dfrac {16\cancel{(t - 10)}(t + 4)\cancel{(t + 5)}} {\cancel{(t + 5)}(t + 6)\cancel{(t - 10)}} $ We are dividing by $t - 10$ , so $t - 10 \neq 0$ Therefore, $t \neq 10$ $a = \dfrac {16(t + 4)} {t + 6} $ $ a = \dfrac{16(t + 4)}{t + 6}; t \neq -5; t \neq 10 $